Download to the marking scheme for all papers

GCE MARKING SCHEME
CHEMISTRY
AS/Advanced
SUMMER 2014
© WJEC CBAC Ltd.
INTRODUCTION
The marking schemes which follow were those used by WJEC for the SUMMER 2014
examination in GCE CHEMISTRY. They were finalised after detailed discussion at
examiners' conferences by all the examiners involved in the assessment. The conferences
were held shortly after the papers were taken so that reference could be made to the full
range of candidates' responses, with photocopied scripts forming the basis of discussion.
The aim of the conferences was to ensure that the marking schemes were interpreted and
applied in the same way by all examiners.
It is hoped that this information will be of assistance to centres but it is recognised at the
same time that, without the benefit of participation in the examiners' conferences, teachers
may have different views on certain matters of detail or interpretation.
WJEC regrets that it cannot enter into any discussion or correspondence about these
marking schemes.
Page
© WJEC CBAC Ltd.
CH1
1
CH2
7
CH4
13
CH5
21
GCE CHEMISTRY - CH1
SUMMER 2014 MARK SCHEME
SECTION A
Q.1
1s22s22p63s23p6
[1]
Q.2
carbon-12 / 12C
[1]
Q.3
any example e.g.
[1]
iron for Haber process / manufacture of ammonia
vanadium(V) oxide in Contact process / manufacture of sulfuric acid
platinum / palladium / rhodium in catalytic converters / to remove toxic gases from
exhaust fumes
nickel in hydrogenation of alkenes / unsaturated oils
Q.4
(a)
Mr = 286.2
(b)
mass = 286.2 × 0.1 = 7.155 / 7.16
4
Q.5
Q.6
Q.7
allow 286
[1]
allow 7.15 / 7.2 based on 286
enthalpy changes = ‒110
234
234
90
91
Th (1)
[1]
[1]
Pa (1) (award 1 mark for 2 correct symbols)
portion to right of Ea1 labelled as molecules that react / shaded
[2]
[1]
Ea2 marked, at lower energy than Ea1, and portion to right labelled as molecules that
react / shaded
[1]
Section A Total [10]
© WJEC CBAC Ltd.
1
SECTION B
Q.8
(a)
same number of protons and electrons (1)
0, 1 and 2 neutrons (1)
(b)
(i)
(ii)
3 energy levels between n = 2 and n = 
becoming closer together
first gap must be < that between n = 1 and n = 2
[1]
any arrow pointing upwards (1)
from n = 1 to n =
(c)
[2]
∞
(1)
[2]
(i)
visible
[1]
(ii)
(not correct because) Balmer series corresponds to energy transitions
involving n = 2 (1)
for ionisation energy need Lyman series / energy transitions involving
n = 1 (1)
[2]
(d)
(i)
Q(g) → Q+(g) + e / accept any symbol
[1]
(ii)
Group 6
[1]
(iii)
In T there is more shielding (1)
The outer electron is further from the nucleus (1)
The increase in shielding outweighs the increase in nuclear
charge / there is less effective nuclear charge (1)
[3]
Legibility of text; accuracy of spelling, punctuation and grammar;
clarity of meaning QWC
[1]
Total [14]
© WJEC CBAC Ltd.
2
Q.9
(a)
(b)
(i)
line drawn that is deflected less by magnetic field
[1]
(ii)
increase strength of the magnetic field
allow decrease charge on charged plates
[1]
(i)
1+ (1)
37
Cl - 37Cl (1)
(ii)
37
Cl2+ (2)
line drawn as m/z 72 (1)
ratio height 6 (1)
(c)
(i)
[2]
allow ½ square tolerance
[2]
% H = 0.84 (1)
C : H : Cl = 10.04 / 12 : 0.84/ 1.01 : 89.12 / 35.5 (1)
= 0.84 : 0.83 : 2.51
= 1 : 1 : 3 empirical formula = CHCl3 (1)
[3]
(ii)
the relative molecular mass / Mr / molar mass
[1]
(iii)
right hand / largest / heaviest m/z peak from mass spectrum
[1]
Total [11]
© WJEC CBAC Ltd.
3
Q.10
(a)
(b)
(a reaction in which) the rate of the forward reaction is equal to the rate
of the backward reaction
[1]
goes darker / more brown (1)
because the (forward) reaction has a +ve H / is endothermic (1)
goes paler / less brown (1)
because there are more moles / molecules on RHS (1)
no change (because catalysts do not affect the position of an equilibrium) (1)
[5]
(c)
(i)
moles N2H4 = 14000/32.04 = 437.0 (1)
this produces 437.0 × 3 = 1311 moles of gas (1)
volume = 1311 × 24 = 3.15 × 104 dm3 (1)
(d)
[minimum 2 sf]
[3]
(ii)
(large volume of) gas produced
[1]
(i)
an acid is a proton / H+ donor
[1]
(ii)
(iii)
→ NO2 + H3O+
[1]
sulfuric acid is behaving as the acid / nitric acid is behaving as a
base (1)
as it donates a proton / as it accepts a proton (1)
[2]
Total [14]
© WJEC CBAC Ltd.
4
Q.11
(a)
(i)
2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l) (state symbols needed)
C(s) allowed as C(gr) or C(graphite)
(b)
[1]
(ii)
(if these elements were reacted together) other products would form/
carbon does not react with hydrogen and oxygen under standard
conditions
[1]
(i)
energy = 100 × 4.2 × 54 =22680
(ii)
moles ethanol = 0.81/46 = 0.0176 (1)
energy change = 22.68
0.0176
[1]
H = ‒1290 (1)
‒ve sign and correct to 3 sf (1)
(c)
[3]
internet value numerically larger (1)
heat losses / incomplete combustion / thermal capacity of calorimeter
ignored (1)
no credit for energy loss
(d)
(e)
[2]
(i)
C3H7OH + 4½O2 → 3CO2 + 4H2O (ignore state symbols)
(ii)
negative enthalpy change means energy in bonds broken is less than
that in bonds made
[1]
(iii)
more bonds broken and made in propanol and therefore more energy
released
[1]
[1]
any 4 from:
both conserve carbon / non-renewable fuel sources / fossil fuels / use
renewable sources
(these gas / liquid) suitable for different uses e.g. ethanol to fuel cars
atom economy gasification is less (some C lost as CO2) / CO2 produced in
gasification is a greenhouse gas
CO is toxic
gasification at high temperature / enzymes need low temperature
enzyme approach therefore saves fuel / gasification needs more energy
[4]
3 max if any reference to destruction of ozone layer
QWC
[2]
The candidate has selected a form and style of writing that is appropriate to
purpose and complexity of the subject matter (1)
Answer has suitable structure (1)
Total [17]
© WJEC CBAC Ltd.
5
Q.12
(a)
to increase rate of reaction / to increase surface area
[1]
(b)
MgCO3 + 2HCl → MgCl2 + CO2 + H2O (ignore state symbols)
(c)
rate starts fast and gradually slows (1)
[1]
because concentration becomes less so fewer collisions (per unit time) /
less frequent collisions / lower probability of collisions (1)
at time = 17/18 min rate = 0 (1)
[3]
(d)
all the solid would all have disappeared / if more carbonate is added further
effervescence is seen
[1]
(e)
(i)
volume CO2 = 200 cm3 (1)
moles CO2 = 200 / 24000 = 0.008333 = moles MgCO3 (1)
[minimum 2 sf]
(ii)
mass MgCO3 = 0.008333 × 84.3 = 0.702 g (1)
% MgCO3 = 0.702 × 100 = 79.0% / 79%
0.889
(e)
[2]
carbon dioxide is soluble in water / reacts with water (1)
volume collected less therefore % / moles of MgCO3 less (1)
(f)
[2]
[2]
use of 40.3 and 84.3 (1)
atom economy = 40.3 / 84.3 × 100 = 47.8% (1)
[2]
Total [14]
Section B Total [70]
© WJEC CBAC Ltd.
6
GCE CHEMISTRY - CH2
SUMMER 2014 MARK SCHEME
SECTION A
Q.1
Van der Waals’ forces < Hydrogen bonds < Covalent bonds
[1]
Q.2
2-methylpentan-1-ol
[1]
Q.3
1 mark showing movement of electrons; 1 mark showing dot and cross of CaCl2 [2]
allow 2-methyl-1-pentanol
-
Cl Ca
Q.4
Q.5
Q.6
Cl
Cl
δ‒ O—Cl δ+
-
Ca2+
Cl
(a)
δ‒ N—H δ+
(b)
Difference in electronegativity is larger in aluminium oxide (so it is ionic) /
the difference is smaller in aluminium chloride (so it is covalent)
[1]
Reagent: Bromine (water)
both for 1 mark
[1]
(1)
Observation(s): hex-2-ene will turn bromine water from orange to colourless,
no change for cyclohexane (1)
[2]
C (1) and E (1) – penalise one mark for each additional incorrect answer
[2]
Section A Total [10]
© WJEC CBAC Ltd.
7
SECTION B
Q.7
(a)
(i)
(ii)
(b)
(i)
(ii)
(iii)
(c)
magnesium
nitrate
barium
chloride
sodium
hydroxide
potassium
carbonate
white
precipitate
white
precipitate
no visible
change
sodium
hydroxide
WHITE
PRECIPITATE
NO VISIBLE
CHANGE
barium chloride
NO VISIBLE
CHANGE
All three correct for 2 marks, two correct for 1 mark
[2]
Name of precipitate: Magnesium carbonate (1)
Ionic equation: Mg2+ + CO32- → MgCO3 (1)
[2]
Sodium hydroxide solution would turn blue/purple
[Ignore references to potassium carbonate]
Potassium carbonate would give a lilac flame
Sodium hydroxide would give a golden yellow flame
Barium chloride would give an apple green flame
(2 for all correct, 1 mark for 2 correct)
1 max if any reference to white flame for magnesium
Barium chloride (1) White precipitate (1)
[1]
[2]
[2]
(i)
Sodium ions surrounded by δ‒ on oxygen atoms of water (1)
Bromide ions surrounded by δ+ on hydrogen atoms of water (1)
Marks can be obtained from a labelled diagram – must show minimum
of two oxygen/hydrogen atoms around sodium/bromide ions
[2]
(ii)
Observation with sodium bromide
cream precipitate (1)
Observation with sodium iodide
yellow precipitate (1)
(iii)
(iv)
[2]
Reagent: (dilute) ammonia solution (1)
Observation with sodium bromide: precipitate dissolves in part
Observations with sodium iodide: precipitate does not change
both observations required for (1)
[If concentrated ammonia (1) used then sodium bromide
will dissolve completely]
[2]
2NaI + Br2 → 2NaBr + I2
[1]
allow ionic equation
Total [16]
© WJEC CBAC Ltd.
8
Q.8
(a)
Boiling temperatures increase with increasing chain length / number of carbon
atoms / relative mass (1)
More carbon atoms leads to greater number of van der Waals’ forces
between molecules (1)
[2]
(b)
(i)
Mass of petroleum gases = 1.2% × 145, 000 = 1740g (1)
Moles of butane = 1740 ÷ 58.1 = 30 mol (1)
Volume of butane = 30 × 24 = 720 dm3 (1)
(ii)
(c)
I.
II.
III.
[3]
ultraviolet light
[1]
Cl2 → 2Cl•
[1]
(Propane forms) propyl radicals / C3H7• (1)
Two C3H7• radicals combine together to make hexane (1) [2]
Brent crude would be better as it has more naphtha (1)
Naphtha is cracked to produce alkenes (1)
Cracking is caused by heating / zeolites / aluminosilicates / porcelain (1)
Any valid equation that produces ethene e.g. C10H22 → C2H4 + C8H18 (1)
Polymerisation: Many small molecules joining together to make a large
molecule (1)
H
n
C
H
H
H
C
C
H
H
H
C
H
n
(1)
Addition polymerisation (1)
e.g. polystyrene, PVC, PTFE and relevant monomer (1)
MAX 6
[6]
QWC: organisation of information clearly and coherently; use of specialist
vocabulary where appropriate
[1]
Total [16]
© WJEC CBAC Ltd.
9
Q.9
(a)
1Fe2O3 + 3CO → 2Fe + 3CO2
(b)
Oxidation state of carbon at start = +2 and at end = +4 so it has been oxidised (1)
Oxidation state of iron at start = +3 and at end = 0 so it has been reduced (1)
Credit 1 mark if all oxidation states are given correctly with incorrect or no
reference to what has been oxidised/reduced
[2]
(c)
(i)
[1]
6:6
[1]
(ii)
O2O2- 90°
O22+
Fe
Diagram must be unambiguous,
either by showing 3 dimensions,
bond angles or through labelling,
must identify iron and oxide as
ions
90°
O2-
O2O2-
[1]
(d)
(e)
Moles FeO = 20,000 ÷ (55.8 +16) = 278.6 mol (1)
Moles Fe = moles FeO = 278.6 mol (1)
Mass Fe = 278.6 × 55.8/1000 = 15.5 kg (1)
Pair of shared electrons in both (1)
Covalent – 1 electron from each atom and
Co-ordinate – 2 electrons from same atom (1)
(f)
[3]
[2]
Lattice / regular arrangement of positive ions (1)
Sea of delocalised electrons (1)
Electrons can move to form an electrical current (1)
Strong forces / bonds between the delocalised electrons and the metal ions
require a lot of energy to break / high temperature to overcome (1)
[4]
QWC: selection of a form and style of writing appropriate to purpose and to
complexity of subject matter
[1]
Total [15]
© WJEC CBAC Ltd.
10
Q.10
(a)
(b)
(i)
Aqueous sodium hydroxide (1) Heat [below 110°C] (1)
[2]
(ii)
Bromobutane cannot form hydrogen bonds (1)
Butan-1-ol can form hydrogen bonds due to its —OH (1)
Hydrogen bonds between butan-1-ol and water molecules allow
butan-1-ol to dissolve (1)
[3]
Acidified dichromate(VI) / acidified manganate(VII) (1)
Heat (1)
[2]
(i)
(ii)
Butanoic acid can form hydrogen bonds between molecules (1)
Bromobutane has van der Waals’ forces between the molecules (1)
Hydrogen bonds are stronger than van der Waals’ so require more
energy to break these (1)
[3]
(iii)
Fractional distillation
[1]
Total [11]
© WJEC CBAC Ltd.
11
Q.11
(a)
(i)
1 mark for arrows in first diagram; 1 mark for arrow in second diagram;
1 mark for all charges
CH3
H
C
C
H
H
H
H
CH3
C
C
H
+
H
H
Br
+
H
H
CH3
C
C
H
Br
-

Br
2 max if incorrect isomer given
[3]
(ii)
(b)
2-bromopropane formed from a secondary carbocation (1)
Secondary carbocations are more stable than primary carbocations (1)
[2]
Empirical formula = C3H5Br
(1)
Molecular formula = C3H5Br
(must show use of mass spectrum to gain this mark) (1)
Two molecular ion peaks as there are two isotopes of bromine
(1)
+
+
Peaks at 15 = CH3 and 41 = C3H5 (1)
550 cm‒1 = C‒Br
1630 cm‒1 = C=C
3030cm‒1 = C‒H
(1)
Molecule is:
Br
CH3
C
C
(1)
H
[6]
H
QWC: legibility of text, accuracy of spelling, punctuation and grammar, clarity
of meaning
[1]
Total [12]
Section B Total [70]
© WJEC CBAC Ltd.
12
H
GCE CHEMISTRY – CH4
SUMMER 2014 MARK SCHEME
SECTION A
Q.1
(a)
(i)
CH3CH2CH2CH2CH3
(ii)
CH3CH2CH2CHCH3
+ Cl2
 CH3CH2CH2CH2CH2Cl
+
HCl

[1]
[1]
(b)
(Anhydrous) aluminium chloride / iron(III) chloride
(c)
(i)
allow AlCl3 / FeCl3
orange / red precipitate
[1]
[1]
(ii)
(1)
—COCH3 groups in any positions
It must contain a C=O group but it is not an aldehyde as it does not react with
Tollens’ reagent
(1)
[2]
(d)
allow KMnO4 / MnO4‒ [1]
(i)
(Alkaline) potassium manganate(VII) (solution)
(ii)
Dilute acid
(iii)
Lithium tetrahydridoaluminate(III) / lithium aluminium hydride
allow LiAlH4
allow HCl / H+
[1]
[1]
(iv)
[1]
(e)
Only the infrared spectrum of benzoic acid would have a peak at 1650–1750 cm‒1 (1)
This is due to the carbonyl group present in the benzoic acid (1)
[2]
Total [12]
© WJEC CBAC Ltd.
13
Q.2
(a)
[1]
(b)
(i)
Acidified potassium dichromate
(ii)
I
allow H+, Cr2O72‒
[1]
An equimolar mixture of two enantiomers / optical isomers
do not accept ‘equal mixture’
[1]
II It has no (apparent) effect on the plane of polarised light
(c)
(d)
(e)
[1]
(i)
But-2-enoic acid; this is because each of the carbon atoms of the double bond
has two different groups / atoms
allow reason based on the other isomer
[1]
(ii)
Any TWO from the following for (1) each
reagent used / temperature / quantities / time of reaction / catalyst / solvent
Reagent(s)
Observation
KOH / I2 or NaOCl / KI (1)
Yellow precipitate (1)
[2]
allow names
[2]
The NMR spectrum will consist of two peaks, as there are two discrete ‘areas’ of
protons; these will be seen at between 2.0 to 2.5 (CH3) and between 2.5 to 3.0 (CH2) (1)
The peak area ratio will be 3:2 for the CH3 and CH2 protons respectively (1)
There will be no splitting of either signal as the protons causing these signals are
not bonded directly to other carbon atoms that also have protons (1)
1 max if only one peak described correctly
[3]
QWC Legibility of text; accuracy of spelling, punctuation and grammar;
clarity of meaning.
[1]
Total [13]
© WJEC CBAC Ltd.
14
Q.3
(a)
(i)
2 mol of ethanol gives 1 mol of ethoxyethane
Moles of ethanol
= 69
46
(1)
= 1.5
 Moles of ethoxyethane if theoretical yield = 0.75
 Moles of ethoxyethane if 45% yield = 0.75 × 0.45 = 0.34
Mass of ethoxyethane = 0.34 × 74 = 25g
(1)
(1) allow error carried forward
[3]
(ii)
Ethene / C2H4
[1]
(iii)
products
(1) for correct curly arrows
(b)
(1) for correct + and ‒
[2]
(iv)
They need to have an N‒H / O‒H / F‒H bond / a highly electronegative atom
bonded to hydrogen
[1]
(i)
For example
[1]
Accept any polybrominated species
Do not accept a monobrominated species
(ii)
(c)
(d)
Bromine decolorised / orange to colourless / white solid
Reagent
Iron(III) chloride solution / FeCl3 (1)
Observation
Purple coloration / solution
(i)
C10H12O1
(1)
[1]
[2]
[1]
(ii)
[1]
© WJEC CBAC Ltd.
15
(e)
Displayed formula, for example
(1)
Functional group
carboxylic acid
(1)
[2]
Total [15]
© WJEC CBAC Ltd.
16
SECTION B
Q.4
(a)
(i)
(Fractional) distillation / (preparative) gas chromatography / HPLC / TLC
column chromatography / solvent extraction
[1]
(ii)
the fragmentation pattern would be different / valid examples given
[1]
(iii)
I
[1]
II Heated electrically / by a naked flame with a water bath (1)
Add compound G to the ethanol until the hot ethanol will (just) not
dissolve any more solute (1)
Filter hot (1)
Allow to cool (1)
Filter (1)
Dry in air / window sill /  60 oC in an oven (1)
[5]
Maximum 4 out of 5 total if second marking point not given
Note 5 marks maximum here
(iv)
I
QWC Information organised clearly and coherently, using specialist
vocabulary where appropriate
[1]
The amine is reacted with sodium nitrite / HCl(aq) or nitrous acid (1)
at a temperature of  10 oC (1)
[2]
II
[1]
© WJEC CBAC Ltd.
17
(b)
(i)
Nucleophilic addition (1)
Accept a mechanism that shows HCN polarisation and nucleophilic addition
as a concerted process
polarisation / charges shown (1) curly arrows on first structure (1)
regeneration of –CN or capture of H+ and curly arrow (1)
(ii)
[4]
Chromophores (1)
The colour will be black (1) as the compound absorbs blue / other colours (1)
[3]
(iii)
[1]
Total [20]
© WJEC CBAC Ltd.
18
Q.5
(a)
C 71.3
 by Ar
 smallest
H 9.6
71.3 = 5.94
12
5.94 = 5
1.193
 O 19.1 (1)
9.6 = 9.6
1.0
19.1 = 1.193
16
9.6 = 8
1.193
1.193 = 1
1.193
(1)
Only one oxygen atom per molecule
 Molecular formula is C5H8O
Silver mirror produced
Ion m/z 29
(1)
present (1)
suggests ethyl group present / CH3CH2
(1)
Structure must be
or
(1)
[6]
(b)
(i)
C11H24
(ii)
Total peak areas
C6H14
+ C3H6
[1]
26 + 13 + 46 = 85
% propene = 13 × 100
85
(iii)
+ C2H4
= 15.(3)
Any THREE points for (1) each
e.g. can it run at a lower temperature (reducing energy costs)
is the yield comparable / better than the yield from the propene process
is the time taken comparable / better than used in the propene process
is there a continued availability of starting materials
can the product be easily / better separated from the reaction mixture
is relatively more expensive equipment needed
is it a batch or continuous process
© WJEC CBAC Ltd.
19
[1]
[3]
(iv)
[1]
(c)
(i)
[1]
(ii)
The production of PTT is an example of condensation polymerisation (1)
The production of poly(propene) is an example of addition polymerisation (1)
Condensation polymerisation needs bifunctional compounds / COOH,OH etc (1)
Addition polymerisation needs a
present in the monomer
Addition polymerisation has an atom economy of 100%
Condensation polymerisation has an atom economy of  100%
(as a co-product is formed)
(1)
(1)
(1) [6]
QWC Selection of a form and style of writing appropriate to purpose and
to complexity of subject matter
[1]
Total [20]
© WJEC CBAC Ltd.
20
GCE CHEMISTRY – CH5
SUMMER 2014 MARK SCHEME
SECTION A
Q.1
(a)
(b)
(i)
NH4+(aq) + OH‒(aq)
NH3(aq) + H2O(l)
Acid 1
Base 2
(1 mark for each pair)
Base 1
Acid 2
[2]
(i)
[NH4+(aq)]/mol dm‒3
[NO2‒(aq)]/mol dm‒3
Initial rate/mol dm‒3 s‒1
1
0.200
0.010
4.00 × 10‒7
2
0.100
0.010
2.00 × 10‒7
3
0.200
0.030
1.20 × 10‒6
4
0.100
0.020
4.00 × 10‒7
(1 mark for each correct answer)
(ii)
[3]
k = 4.00 × 10‒7 = 2.0 × 10‒4
0.200 × 0.010
(1)
Units = mol‒1 dm3 s‒1
(1)
(iii)
No change
(iv)
Increases
[2]
[1]
If temperature is increased rate increases (1)
and since concentrations do not change the rate constant must increase
(or similar) (1)
[2]
Total [10]
© WJEC CBAC Ltd.
21
Q.2
(a)
(b)
Kw = [H+][OH‒]
Units = mol2 dm-6
(i)
(ii)
(1)
(1)
In pure water [H+] = [OH‒] or [H+] = √1.0 × 10‒14
(1)
pH = ‒log 10‒7 = 7
(1)
moles acid = 0.1 × 10 = 0.001
1000
pH = ‒log 0.001 = 3
(1)
(1)
1.78 × 10‒5 = [H+] × 0.02
0.01
(1)
[H+] = 8.90 × 10‒6
(1)
pH = 5.05
(d)
[2]
Final volume of solution is 1000 cm3 so acid has been diluted by a
factor of 100 so final concentration of acid is 0.001
or
(c)
[2]
allow 5 or 5.1
[2]
(1)
The solution is a buffer
[3]
(1)
Solution contains a large amount of CH3COOH and CH3COO‒ ions
(Accept correct equations)
(1)
‒
+
When an acid is added, the CH3COO ions react with the H ions, removing
them from solution and keeping the pH constant
(1)
[3]
Total [12]
© WJEC CBAC Ltd.
22
Q.3
(a)
.. ..
H:O:O:H
..
(b)
(c)
[1]
..
20 dm3 oxygen = 0.83 mol
(1)
Moles H2O2 = 1.67 and [H2O2] = 1.67 mol dm‒3
(1)
(i)
(ii)
[2]
Variable oxidation states / partially filled 3d energy levels /ability to
adsorb ‘molecules’ / form complexes (or temporary bonds) with
reacting molecules
(Accept any two answers)
Do not accept ‘empty / unfilled d-orbitals’
[2]
3d orbitals split by ligands
(1)
Three d-orbitals have lower energy, two have higher energy (1)
Electrons absorb (visible light) energy to jump from lower level to
higher level (1)
The colour is that due to the remaining / non-absorbed frequencies (1)
(Appropriate diagrams are acceptable alternatives)
[4]
(d)
QWC Legibility of text; accuracy of spelling, punctuation and
grammar, clarity of meaning
[1]
(i)
MnO4‒ + 8H+ + 5e‒
Mn2+ + 4H2O
[1]
(ii)
5H2O2 + 6H+ + 2MnO4‒
2Mn2+ + 5O2 + 8H2O
[2]
(Mark consequentially from (i) – 1 mark if formulae correct but
equation not balanced properly)
(iii)
(e)
Moles MnO4‒ = 0.02 × 14.8 = 2.96 × 10‒4
1000
(1)
Moles H2O2 = 7.40 × 10‒4
(1)
Concentration H2O2 = 7.40 × 10‒4 = 0.037 mol dm‒3
0.020
(1)
Oxidation state of oxygen starts at ‒1 (in peroxide)
Oxidation state in water is ‒2 (reduced)
oxidation state in oxygen is 0 (oxidised)
[3]
(1)
(1)
[2]
Total [18]
© WJEC CBAC Ltd.
23
SECTION B
Q.4
(a)
(b)
(i)
Oxidising agent
(ii)
A = lead(II) chloride / PbCl2
(1)
B = chlorine / Cl2
(1)
[2]
(iii)
[Pb(OH)6]4‒ / [Pb(OH)4]2‒ / Na4[Pb(OH)6] etc.
[1]
(iv)
Yellow
[1]
(v)
PbO + 2HNO3
(i)
[1]
Pb(NO3)2 + H2O
[1]
Each C atom covalently bonded to three other C atoms forming layers
(1)
Layers held together by weak intermolecular forces
(1)
BN is isoelectronic with C so it forms similar structures
(1)
Graphite conducts electricity since electrons are delocalised but in BN,
each N has a full unbonded p-orbital and each B has an empty
unbonded p-orbital so it does not conduct electricity
(1)
[4]
(Accept electrons are not delocalised in BN so it does not conduct
electricity)
QWC The information is organised clearly and coherently, using
specialist vocabulary where appropriate
[1]
(c)
(ii)
Wear-resistant coatings/catalyst support/for mounting high power
electronic components / drills in industry / cutting instruments
[1]
(i)
ΔG = ΔH ‒ T ΔS
(ii)
(ΔG = 0 for reaction to be spontaneous)
T = 1.92
0.0067
(1)
T = 286.6 K
(1)
(1)
[3]
Changes in temperature (above or below 286.6 K) caused the tin to
change form making it unstable (and causing it to disintegrate)
[1]
© WJEC CBAC Ltd.
24
(d)
(i)
(At the anode)
2H+ + 2e‒
H2
+
‒
(At the cathode)
O2 + 4H + 4e
(Overall reaction)
2H2 + O2
(1)
2H2O
2H2O
(1)
(1)
[3]
(ii)
Hydrogen is difficult to store / takes up large volume / too flammable /
explosive / produced from fossil fuels which leads to a net energy loss
/ Pt electrodes very expensive
[1]
Total [20]
© WJEC CBAC Ltd.
25
Q.5
(a)
(i)
Cold
Cl2 + 2NaOH
NaCl + NaClO + H2O
Warm 3Cl2 + 6NaOH
(1)
5NaCl + NaClO3 + 3H2O (1)
[2]
(ii)
Disproportionation
[1]
(b)
P can (extend the normal octet of electrons) by using 3d orbitals /
P can promote 3s electron to 3d orbital
(1)
N cannot do this since it is in the second period / 3d orbitals not available (1)
[2]
(c)
The terms involved are: lattice breaking enthalpy which is endothermic
(1)
and hydration enthalpy which is exothermic
(1)
ΔH solution = ΔH lattice breaking + ΔH hydration (or similar)
(1)
If ΔH solution is negative then the ionic solid will be soluble
(1)
[4]
QWC Selection of a form and style of writing appropriate to purpose and to
complexity of subject matter
[1]
(d)
(i)
(ii)
(e)
(i)
Iodide
(1)
Only one with less positive standard potential than
Fe3+, Fe2+ half-cell
(1)
nd
(2 mark can be obtained from calculation value and statement)
Pt(s) Fe2+(aq), Fe3+(aq) Ce4+(aq), Ce3+(aq) Pt (s)
(1)
EMF = 1.45 – 0.77 = 0 .68 V
(1)
Kc = [CH3COOCH3][H2O]
[CH3COOH][CH3OH]
(1)
No units
(1)
(ii)
moles = 1.25 × 32.0 = 0.04(0)
1000
(iii)
[CH3COOH] = 0.04, therefore 0.06 used in reaction and
[2]
[2]
[2]
[1]
[CH3COOCH3] = 0.06, [H2O] = 0.06 and
(iv)
[CH3OH] = 0.083 – 0.06 = 0.023
(1)
Kc = 0.06 × 0.06 = 3.91
0.04 × 0.023
(1)
[2]
Value of Kc decreases since the equilibrium shifts to the left /
the forward reaction is exothermic
[1]
Total [20]
© WJEC CBAC Ltd.
26
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Cardiff CF5 2YX
Tel No 029 2026 5000
Fax 029 2057 5994
E-mail: [email protected]
website: www.wjec.co.uk
© WJEC CBAC Ltd.